If $p$ is the length of perpendicular from the origin on the line $\frac{x}{a}+\frac{y}{b}=1$ and $a^2, p^2$ and $b^2$ are in AP, the show that $a^4+b^4=0$.
Given equation of line is,
$$\frac{x}{a}+\frac{y}{b}=1\quad \text{.... (i)}$$
Perpendicular length from the origin on the line (i) is given by $p$
$$\begin{aligned} \text{i.e.,}\quad & p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{a b}{\sqrt{a^2+b^2}} \\ \therefore \quad & p^2=\frac{a^2 b^2}{a^2+b^2} \end{aligned}$$
Given that, $a^2, p^2$ and $b^2$ are in AP.
$$ \begin{array}{lr} \therefore & 2 p^2=a^2+b^2 \\ \Rightarrow & \frac{2 a^2 b^2}{a^2+b^2}=a^2+b^2 \\ \Rightarrow & 2 a^2 b^2=\left(a^2+b^2\right)^2 \\ \Rightarrow & 2 a^2+b^2=a^4+b^4+2 a^2 b^2 \\ \Rightarrow & a^4+b^4=0 \end{array}$$
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